Chapter 17.1, Problem 4PP

Solution

Interpretation:

- The equilibrium constant expression for the reaction of solid iron with chlorine gas to form solid iron $III$ chloride $(FeC{l}_3)$ is to be written.

Concept introduction:

- The equilibrium constant expression of the reaction is the ratio of molar concentration of reactant solid iron and chlorine gas and molar concentration of product − Ferric chloride $(FeC{l}_3).$

Balanced chemical equation for this reaction is the symbolic representation is the symbolic representation of this reaction.

$2Fe(s)+3C{l}_2(g)\to 2FeC{l}_3(s)$

$\text{Iron solid + chlorine gas}\to \text{iron chloride}$

Equilibrium constant expression for this reaction

$K_C=\frac{{[FeC{l}_3(s)]}^2}{{[Fe(s)]}^2{[Cl(g)]}^3}$

$\because [FeC{l}_3(s)]$ and ${[Fe(s)]}^2$ is solid.

So,

$K_C=\frac{1}{{[C{l}_2(g)]}^3}$.

The iron reach with chlorine rigorously to form a cloud of iron $III$ chloride.

Chemical equation for this reaction

$2Fe(s)+3C{l}_2(g)\to 2FeC{l}_3(s)$

It is a heterogenous reaction because the reactant in different state. Iron is solid and chlorine is in gaseous state.

Equilibrium constant expression $\text{'}{K}_C\text{'}$ for this reaction is

$K_C=\frac{\text{[Product]}}{\text{[Reactant]}}$

$K_C=\frac{{[FeC{l}_3(g)]}^2}{{[Fe(s)]}^2{[Cl(g)]}^3}$

$\because [FeC{l}_3(s)]$ and ${[Fe(s)]}^2$ is solid so they are omitted form the value of $\text{'}{K}_C\text{'}$

$\therefore K_C=\frac{1}{{[C{l}_2(g)]}^3}$

The chemical equation is as follows

$2Fe(s)+3C{l}_2(g)\to 2FeC{l}_3(s)$

Equilibrium constant for this

$K_C=\frac{1}{{[C{l}_2(g)]}^3}$